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          <h1 class="post-title" itemprop="name headline">[机器学习Lesson 2]代价函数之线性回归算法</h1>
        

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<p>本章内容主要是介绍：单变量线性回归算法（Linear regression with one variable）</p>
</blockquote>
<h2 id="1-线性回归算法（linear-regression）"><a href="#1-线性回归算法（linear-regression）" class="headerlink" title="1. 线性回归算法（linear regression）"></a>1. 线性回归算法（linear regression）</h2><h3 id="1-1-预测房屋价格"><a href="#1-1-预测房屋价格" class="headerlink" title="1.1 预测房屋价格"></a>1.1 预测房屋价格</h3><p>下图是俄勒冈州波特兰市的住房价格和面积大小的关系：<br><a id="more"></a></p>
<p><img src="https://s1.ax2x.com/2018/03/25/15HwS.png" alt="数据集包含俄勒冈州波特兰市的住房价格"></p>
<p>该问题属于监督学习中的回归问题，让我们来复习一下：</p>
<ul>
<li>监督学习（Supervised’Learning’）:对示例数据给出“正确答案”。<ul>
<li>回归问题（Regression ‘Problem’）:根据之前的数据预测出一个准确的输出值 。</li>
</ul>
</li>
</ul>
<h3 id="1-2-训练集"><a href="#1-2-训练集" class="headerlink" title="1.2 训练集"></a>1.2 训练集</h3><p><img src="https://s1.ax2x.com/2018/03/25/15bGR.png" alt="image"></p>
<ul>
<li><strong>m</strong>=训练样本数量</li>
<li><strong>x’s</strong>=输入变量/特征量</li>
<li><strong>y’s</strong>=输出变量/目标变量，预测结果</li>
</ul>
<p>（x,y）表示一个训练样本。</p>
<p>x(1) 指的是 第一个训练集里值为2104的输入值， 这个就是第一行里的x x(2) 等于1416。这是第二个x y(1) 等于460，这是第一个训练集样本的y值， 这就是(1)所代表的含义。</p>
<p>这就是一个监督学习算法的工作方式，我们可以看到这里有我们的训练集里房屋价格，我们把它喂给我们的学习算法，然后输出一个函数。  </p>
<p>按照惯例，通常表示为小写h代表hypothesis(假设) h表示一个函数。输入是房屋尺寸大小，就像你朋友想出售的房屋。因此，h 根据输入的 x 值来得出 y 值。 y值对应房子的价格。所以h是一个从x到y的函数映射 。</p>
<ul>
<li>y关于x的线性函数 ：</li>
</ul>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">hθ(x)=θ0+θ1*x</span><br></pre></td></tr></table></figure>
<p><img src="https://s1.ax2x.com/2018/03/27/1Uj5p.png" alt="image"></p>
<p>这个模型被称为<code>线性回归(linear regression)模型</code>。 这实际上是关于单个变量的线性回归，这个变量就是x 根据x来预测所有的价格函数。同时， 对于这种模型有另外一个名称，称作<code>单变量线性回归</code> 单变量是对一个变量的一种特别的表述方式。总而言之 这就是线性回归。</p>
<h2 id="2-代价函数-Cost-Function"><a href="#2-代价函数-Cost-Function" class="headerlink" title="2. 代价函数(Cost Function)"></a>2. 代价函数(Cost Function)</h2><p>任何能够衡量模型预测出来的值h(θ)与真实值y之间的差异的函数都可以叫做代价函数C(θ)，如果有多个样本，则可以将所有代价函数的取值求均值，记做J(θ)。</p>
<p><code>J(θ0,θ1)=12m$\sum$i=1m(y^i−yi)2=12m∑i=1m(hθ(xi)−yi)2</code></p>
<p><img src="https://s1.ax2x.com/2018/03/27/1Um0y.md.png" alt="image"></p>
<ul>
<li><p>m：训练样本的个数；</p>
</li>
<li><p>hθ(x)：用参数θ和x预测出来的y值；</p>
</li>
<li><p>y：原训练样本中的y值，也就是标准答案</p>
</li>
<li><p>上角标(i)：第i个样本</p>
</li>
</ul>
<h2 id="3-代价函数1-简化版-当θ0-0时"><a href="#3-代价函数1-简化版-当θ0-0时" class="headerlink" title="3. 代价函数1(简化版):当θ0=0时"></a>3. 代价函数1(简化版):当θ0=0时</h2><p>hθ(x)=θ1x，如下图：</p>
<h5 id="重要公式"><a href="#重要公式" class="headerlink" title="重要公式"></a>重要公式</h5><p><img src="https://s1.ax2x.com/2018/03/26/1aftG.md.png" alt="image"></p>
<ul>
<li>Hypothesis: 假设。这个例子中是尺寸对于房价关系的预测。</li>
<li>Parameters: 参数。</li>
<li>Cost Function：代价函数。</li>
<li>Goal: 优化目标。代价最小化。</li>
</ul>
<h3 id="3-1-斜率为1时的代价函数"><a href="#3-1-斜率为1时的代价函数" class="headerlink" title="3.1 斜率为1时的代价函数"></a>3.1 斜率为1时的代价函数</h3><p><img src="https://s1.ax2x.com/2018/03/26/1aoPz.png" alt="image"></p>
<h5 id="（1）假设函数"><a href="#（1）假设函数" class="headerlink" title="（1）假设函数"></a>（1）假设函数</h5><blockquote>
<p>x轴为面积，y轴为房价  </p>
</blockquote>
<p>假设函数 h(x) 对于一个固定的θ1，这是一个关于x 的函数。 所以这个假设函数就是一个关于 x 这个房子大小的函数。</p>
<h5 id="（2）代价函数"><a href="#（2）代价函数" class="headerlink" title="（2）代价函数"></a>（2）代价函数</h5><blockquote>
<p>x轴为假设函数的斜率，y即代价大小</p>
</blockquote>
<p>代价函数 J 是一个关于参数 θ1 的函数，而 θ1 控制着这条直线的斜率 。</p>
<h3 id="3-2-斜率为0-5时的代价函数"><a href="#3-2-斜率为0-5时的代价函数" class="headerlink" title="3.2 斜率为0.5时的代价函数"></a>3.2 斜率为0.5时的代价函数</h3><p><img src="https://s1.ax2x.com/2018/03/26/1aiwn.png" alt="image"></p>
<p>斜率为0.5时，取3个样本（m=3）：（0.5，1），（1，2），（1.5，3）。套公式得出J(0.5)=0.58<br>同理，J(0)=1/6(1²+2²+3²)=14/6，求出更多的点之后，我们得出类似以下函数：</p>
<p><img src="https://s1.ax2x.com/2018/03/27/1en09.png" alt="image"></p>
<p>学习算法的优化目标是我们想找到一个 θ1 的值，来将 J(θ1) 最小化。这是我们线性回归的目标函数。 上面的曲线中，让 J(θ1) 最小化的值是 θ1=1。这个确实就对应着最佳的通过了数据点的拟合直线 。这条直线就是由 θ1=1 的设定而得到的。 对于这个特定的训练样本，我们最后能够完美地拟合 这就是为什么最小化 J(θ1)，对应着寻找一个最佳拟合直线的目标。</p>
<h2 id="4-代价函数2-完整版"><a href="#4-代价函数2-完整版" class="headerlink" title="4. 代价函数2:完整版"></a>4. 代价函数2:完整版</h2><p>包含θ0、θ1两个参数的代价函数呈现出来的是类似下图的三维曲面图，两个轴分别表示θ0、θ1。</p>
<p><img src="https://s1.ax2x.com/2018/03/26/1rDX6.png" alt="image"></p>
<p>在ML中，一般使用轮廓图( contour plot 或 contour figure 的意思)描述该模型。</p>
<h3 id="4-1-轮廓图简介"><a href="#4-1-轮廓图简介" class="headerlink" title="4.1 轮廓图简介"></a>4.1 轮廓图简介</h3><p><img src="https://s1.ax2x.com/2018/03/26/1r7za.png" alt="image"></p>
<p>右侧图形就是一个轮廓图，两个轴分别表示θ0和θ1。 而这些一圈一圈的椭圆形，每一个圈就表示J(θ0,θ1) 相同的所有点的集合。</p>
<p>如图选取三个点，这三个点都表示相同的 J(θ0,θ1) 的值。横纵坐标分别是θ0, θ1 这三个点的 J(θ0,θ1) 值是相同的。我们需要算的代价函数即为圆心的点，此时我们的代价最小。</p>
<h3 id="4-2-第一组数据"><a href="#4-2-第一组数据" class="headerlink" title="4.2 第一组数据"></a>4.2 第一组数据</h3><p> 我们选取一组数据，<code>θ0=800</code>，<code>θ1=-0.15</code>，此时我们可以对应得到一个左边这样一条线。</p>
<p> <img src="https://s1.ax2x.com/2018/03/26/1rdmz.png" alt="image"></p>
<p>以这组 θ0，θ1 为参数的这个假设 h(x) 并不是数据的较好拟合。并且你也发现了这个代价值 距离最小值点还很远。也就是说这个代价值还是算比较大的，因此不能很好拟合数据。</p>
<h3 id="4-3-第二组数据"><a href="#4-3-第二组数据" class="headerlink" title="4.3 第二组数据"></a>4.3 第二组数据</h3><p><img src="https://s1.ax2x.com/2018/03/27/1eetp.png" alt="选取第二组数据"></p>
<p><code>θ0=360</code>，<code>θ1=0</code>。我们可以得到<code>h(x)=360+0*x</code>这样一条直线。同样不能很好的拟合数据。</p>
<h3 id="4-4-第三组数据"><a href="#4-4-第三组数据" class="headerlink" title="4.4 第三组数据"></a>4.4 第三组数据</h3><p>最后一个例子：<br><img src="https://s1.ax2x.com/2018/03/27/1ex3h.md.png" alt="image"></p>
<p>这个点其实不是最小值，但已经非常靠近最小值点了。 这个点对数据的拟合就很不错，它对应这样两个θ0 和 θ1 的值。同时也对应这样一个 h(x) 这个点虽然不在最小值点，但非常接近了。 因此误差平方和，或者说 训练样本和假设的距离的平方和，这个距离值的平方和 非常接近于最小值，尽管它还不是最小值。</p>
<h2 id="5-小结"><a href="#5-小结" class="headerlink" title="5. 小结"></a>5. 小结</h2><p>通过这些图形，本篇文章主要是帮助理解这些代价函数 J 所表达的值；它们是什么样的它们对应的假设是什么样的；以及什么样的假设对应的点更接近于代价函数J的最小值。</p>
<p>我们真正需要的是一种有效的算法，能够自动地找出这些使代价函数J取最小值的参数θ0和θ1来。我们也不希望编个程序 把这些点画出来，然后人工的方法来读出这些点的数值，这很明显不是一个好办法。  </p>
<p>事实上在深入机器学习的过程中， 我们会遇到更复杂、更高维度、更多参数的情况。而这些情况是很难画出图的，因此更无法将其可视化，因此我们真正需要的，是编写程序来找出这些最小化代价函数的θ0和θ1的值。在后续文章中将介绍一种算法 能够自动地找出能使代价函数 J最小化的参数θ0和θ1的值。</p>
<hr>
<p>本文资料部分来源于吴恩达 (Andrew Ng) 博士的斯坦福大学机器学习公开课视频教程。   </p>
<p>[1]网易云课堂机器学习课程：<br><a href="http://open.163.com/special/opencourse/machinelearning.html" target="_blank" rel="noopener">http://open.163.com/special/opencourse/machinelearning.html</a><br>[2]coursera课程：<br><a href="https://www.coursera.org/learn/machine-learning/" target="_blank" rel="noopener">https://www.coursera.org/learn/machine-learning/</a></p>

      
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              <div class="post-toc-content"><ol class="nav"><li class="nav-item nav-level-2"><a class="nav-link" href="#1-线性回归算法（linear-regression）"><span class="nav-text">1. 线性回归算法（linear regression）</span></a><ol class="nav-child"><li class="nav-item nav-level-3"><a class="nav-link" href="#1-1-预测房屋价格"><span class="nav-text">1.1 预测房屋价格</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#1-2-训练集"><span class="nav-text">1.2 训练集</span></a></li></ol></li><li class="nav-item nav-level-2"><a class="nav-link" href="#2-代价函数-Cost-Function"><span class="nav-text">2. 代价函数(Cost Function)</span></a></li><li class="nav-item nav-level-2"><a class="nav-link" href="#3-代价函数1-简化版-当θ0-0时"><span class="nav-text">3. 代价函数1(简化版):当θ0=0时</span></a><ol class="nav-child"><li class="nav-item nav-level-5"><a class="nav-link" href="#重要公式"><span class="nav-text">重要公式</span></a></li></ol></li></ol><li class="nav-item nav-level-3"><a class="nav-link" href="#3-1-斜率为1时的代价函数"><span class="nav-text">3.1 斜率为1时的代价函数</span></a><ol class="nav-child"><li class="nav-item nav-level-5"><a class="nav-link" href="#（1）假设函数"><span class="nav-text">（1）假设函数</span></a></li><li class="nav-item nav-level-5"><a class="nav-link" href="#（2）代价函数"><span class="nav-text">（2）代价函数</span></a></li></ol></li><li class="nav-item nav-level-3"><a class="nav-link" href="#3-2-斜率为0-5时的代价函数"><span class="nav-text">3.2 斜率为0.5时的代价函数</span></a></li><li class="nav-item nav-level-2"><a class="nav-link" href="#4-代价函数2-完整版"><span class="nav-text">4. 代价函数2:完整版</span></a><ol class="nav-child"><li class="nav-item nav-level-3"><a class="nav-link" href="#4-1-轮廓图简介"><span class="nav-text">4.1 轮廓图简介</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#4-2-第一组数据"><span class="nav-text">4.2 第一组数据</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#4-3-第二组数据"><span class="nav-text">4.3 第二组数据</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#4-4-第三组数据"><span class="nav-text">4.4 第三组数据</span></a></li></ol></li><li class="nav-item nav-level-2"><a class="nav-link" href="#5-小结"><span class="nav-text">5. 小结</span></a></li></div>
            

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                    };
                  }

                  var slicesOfTitle = [];
                  if (indexOfTitle.length != 0) {
                    slicesOfTitle.push(mergeIntoSlice(title, 0, title.length, indexOfTitle));
                  }

                  var slicesOfContent = [];
                  while (indexOfContent.length != 0) {
                    var item = indexOfContent[indexOfContent.length - 1];
                    var position = item.position;
                    var word = item.word;
                    // cut out 100 characters
                    var start = position - 20;
                    var end = position + 80;
                    if(start < 0){
                      start = 0;
                    }
                    if (end < position + word.length) {
                      end = position + word.length;
                    }
                    if(end > content.length){
                      end = content.length;
                    }
                    slicesOfContent.push(mergeIntoSlice(content, start, end, indexOfContent));
                  }

                  // sort slices in content by search text's count and hits' count

                  slicesOfContent.sort(function (sliceLeft, sliceRight) {
                    if (sliceLeft.searchTextCount !== sliceRight.searchTextCount) {
                      return sliceRight.searchTextCount - sliceLeft.searchTextCount;
                    } else if (sliceLeft.hits.length !== sliceRight.hits.length) {
                      return sliceRight.hits.length - sliceLeft.hits.length;
                    } else {
                      return sliceLeft.start - sliceRight.start;
                    }
                  });

                  // select top N slices in content

                  var upperBound = parseInt('1');
                  if (upperBound >= 0) {
                    slicesOfContent = slicesOfContent.slice(0, upperBound);
                  }

                  // highlight title and content

                  function highlightKeyword(text, slice) {
                    var result = '';
                    var prevEnd = slice.start;
                    slice.hits.forEach(function (hit) {
                      result += text.substring(prevEnd, hit.position);
                      var end = hit.position + hit.length;
                      result += '<b class="search-keyword">' + text.substring(hit.position, end) + '</b>';
                      prevEnd = end;
                    });
                    result += text.substring(prevEnd, slice.end);
                    return result;
                  }

                  var resultItem = '';

                  if (slicesOfTitle.length != 0) {
                    resultItem += "<li><a href='" + articleUrl + "' class='search-result-title'>" + highlightKeyword(title, slicesOfTitle[0]) + "</a>";
                  } else {
                    resultItem += "<li><a href='" + articleUrl + "' class='search-result-title'>" + title + "</a>";
                  }

                  slicesOfContent.forEach(function (slice) {
                    resultItem += "<a href='" + articleUrl + "'>" +
                      "<p class=\"search-result\">" + highlightKeyword(content, slice) +
                      "...</p>" + "</a>";
                  });

                  resultItem += "</li>";
                  resultItems.push({
                    item: resultItem,
                    searchTextCount: searchTextCount,
                    hitCount: hitCount,
                    id: resultItems.length
                  });
                }
              })
            };
            if (keywords.length === 1 && keywords[0] === "") {
              resultContent.innerHTML = '<div id="no-result"><i class="fa fa-search fa-5x" /></div>'
            } else if (resultItems.length === 0) {
              resultContent.innerHTML = '<div id="no-result"><i class="fa fa-frown-o fa-5x" /></div>'
            } else {
              resultItems.sort(function (resultLeft, resultRight) {
                if (resultLeft.searchTextCount !== resultRight.searchTextCount) {
                  return resultRight.searchTextCount - resultLeft.searchTextCount;
                } else if (resultLeft.hitCount !== resultRight.hitCount) {
                  return resultRight.hitCount - resultLeft.hitCount;
                } else {
                  return resultRight.id - resultLeft.id;
                }
              });
              var searchResultList = '<ul class=\"search-result-list\">';
              resultItems.forEach(function (result) {
                searchResultList += result.item;
              })
              searchResultList += "</ul>";
              resultContent.innerHTML = searchResultList;
            }
          }

          if ('auto' === 'auto') {
            input.addEventListener('input', inputEventFunction);
          } else {
            $('.search-icon').click(inputEventFunction);
            input.addEventListener('keypress', function (event) {
              if (event.keyCode === 13) {
                inputEventFunction();
              }
            });
          }

          // remove loading animation
          $(".local-search-pop-overlay").remove();
          $('body').css('overflow', '');

          proceedsearch();
        }
      });
    }

    // handle and trigger popup window;
    $('.popup-trigger').click(function(e) {
      e.stopPropagation();
      if (isfetched === false) {
        searchFunc(path, 'local-search-input', 'local-search-result');
      } else {
        proceedsearch();
      };
    });

    $('.popup-btn-close').click(onPopupClose);
    $('.popup').click(function(e){
      e.stopPropagation();
    });
    $(document).on('keyup', function (event) {
      var shouldDismissSearchPopup = event.which === 27 &&
        $('.search-popup').is(':visible');
      if (shouldDismissSearchPopup) {
        onPopupClose();
      }
    });
  </script>





  

  
  <script src="https://cdn1.lncld.net/static/js/av-core-mini-0.6.4.js"></script>
  <script>AV.initialize("1ksH739lNLQGmPbiKV7caYHV-gzGzoHsz", "kgyOnl48BVfVTzUF8NaU6gFY");</script>
  <script>
    function showTime(Counter) {
      var query = new AV.Query(Counter);
      var entries = [];
      var $visitors = $(".leancloud_visitors");

      $visitors.each(function () {
        entries.push( $(this).attr("id").trim() );
      });

      query.containedIn('url', entries);
      query.find()
        .done(function (results) {
          var COUNT_CONTAINER_REF = '.leancloud-visitors-count';

          if (results.length === 0) {
            $visitors.find(COUNT_CONTAINER_REF).text(0);
            return;
          }

          for (var i = 0; i < results.length; i++) {
            var item = results[i];
            var url = item.get('url');
            var time = item.get('time');
            var element = document.getElementById(url);

            $(element).find(COUNT_CONTAINER_REF).text(time);
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          for(var i = 0; i < entries.length; i++) {
            var url = entries[i];
            var element = document.getElementById(url);
            var countSpan = $(element).find(COUNT_CONTAINER_REF);
            if( countSpan.text() == '') {
              countSpan.text(0);
            }
          }
        })
        .fail(function (object, error) {
          console.log("Error: " + error.code + " " + error.message);
        });
    }

    function addCount(Counter) {
      var $visitors = $(".leancloud_visitors");
      var url = $visitors.attr('id').trim();
      var title = $visitors.attr('data-flag-title').trim();
      var query = new AV.Query(Counter);

      query.equalTo("url", url);
      query.find({
        success: function(results) {
          if (results.length > 0) {
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            counter.fetchWhenSave(true);
            counter.increment("time");
            counter.save(null, {
              success: function(counter) {
                var $element = $(document.getElementById(url));
                $element.find('.leancloud-visitors-count').text(counter.get('time'));
              },
              error: function(counter, error) {
                console.log('Failed to save Visitor num, with error message: ' + error.message);
              }
            });
          } else {
            var newcounter = new Counter();
            /* Set ACL */
            var acl = new AV.ACL();
            acl.setPublicReadAccess(true);
            acl.setPublicWriteAccess(true);
            newcounter.setACL(acl);
            /* End Set ACL */
            newcounter.set("title", title);
            newcounter.set("url", url);
            newcounter.set("time", 1);
            newcounter.save(null, {
              success: function(newcounter) {
                var $element = $(document.getElementById(url));
                $element.find('.leancloud-visitors-count').text(newcounter.get('time'));
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                console.log('Failed to create');
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    $(function() {
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      if ($('.leancloud_visitors').length == 1) {
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        showTime(Counter);
      }
    });
  </script>



  

  

  
  

  

  

  

</body>
</html>
